DC Power Supply
DC power supply is a significant element in most electronics and electrical work. Its main function is to supply a desired value of direct current to a specific load. It is imperative that the appropriate amount of DC power is supplied to electrical load for correct functioning without damage. As a consequence, DC power supply is significant in supplying the desired amount of direct current power. In this project, a 5 V DC switching power supply will be designed. The power supply will be designed to drive a 50 ohm load with less than 5 percent ripple and less than 5 percent error.
Design
The basic components for a DC power supply include a step down transformer, full wave rectifier, filter, voltage regulator and a load. The value of the components for this design will be established based on the outlined requirements.
Theoretical Results
The theoretical results expected from the design of a 5 V DC switching power supply capable of driving a 50 ohm load with less than 5 percent ripple and less than 5 percent error should follow a certain expectation. The required values of the DV power supply should produce specific theoretical values according to the selected design components of the power supply. The components selected for this design include a full bridge rectifier consisting of four diodes, a step down transformer, diode rectifier, filter, voltage regulator and load.
It is imperative to determine the theoretical outcome of the resistance of the system. It should be able to support a 5 ohm load. The resistance of the system is obtained by attaining the quotient of the resultant voltage across the system and the sum of the minimum and the maximum current (Pizzi & O'meara,1993).
R= (Vcmin – Vzo – RzIzmin) / (Izmin + I Lmax)
Where; Vzo is the voltage across the zener diode (5.1 V), Rz is given as the 7 ohms, Izmin is the minimum current fro the zener diode (5 mA) , the current through the transfer is given by ILmax (30mA).
The minimum voltage Vc min is obtained by
Vc min = 17v (rms value) - .7(diode) – .5(5% ripples) = 15.77V
As a result, the R for the system is obtained as 303 Ohms.
The ripple voltage is obtained by
Vr= Vp / 2*f*C*R
Where Vp is the peak voltage, f is the frequency, C is the capacitance and R is the resistance
Vp/R = [((16.77+15.7)/2 ) -5] / 303 = 37mA
Therefore
C = 37mA/2*(60)*.5
C= 61uF
Ripples equations
Ripple = #ripple division (200mV/ 5 divisions)*.5
Error%:
Error% = (Voltage – 5V)/5V
Testing
The designed system was tested through measuring the desired output such as the resistor, percentage ripple, the gain and the percentage error. A series of data was tested and recorded at different resistance of the system.
Conclusion
The measured results deviated slightly from the expected theoretical results. The slight deviation may be due to human error as well as the selected design components. The 5 V DC switching power supply was expected to drive a 50 ohm load with less than 5 percent ripple and less than 5 percent error. According to the measured results, at 300 ohm the power supply reported a gain of 3.2 V, percentage ripple of 2 and 36 percent percentage error.
References
Pizzi, R. J., & O'meara, J. M. (1993). U.S. Patent No. 5,258,701. Washington, DC: U.S. Patent and Trademark Office.
